Question

$\underset{0}{\overset{1}{\int }}\left(x{e}^x+\cos \frac{\mathrm{\pi x}}{4}\right)dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^1\left(x{e}^x+\cos \frac{\pi x}{4}\right)\mathit{d}x\mathit{.}\mathit{}\mathrm{Then}, \\ I={\int }_0^{1}x{e}^{x}dx+{\int }_0^1\cos \frac{\pi x}{4}dx \\ \mathrm{Integrating}\mathrm{first}\mathrm{term}\mathrm{by}\mathrm{parts} \\ I=\left\{{\left[x{e}^x\right]}_0^1-{\int }_0^{1}1e^{x}dx\right\}+{\left[\frac{\sin {\displaystyle \frac{\pi x}{4}}}{{\displaystyle \frac{\mathrm{\pi }}{4}}}\right]}_0^1 \\ \Rightarrow I={\left[x{e}^x\right]}_0^1-{\left[e^x\right]}_0^1+{\left[\frac{\sin {\displaystyle \frac{\pi x}{4}}}{{\displaystyle \frac{\mathrm{\pi }}{4}}}\right]}_0^1 \\ \Rightarrow I=e-e+1+\frac{4}{\mathrm{\pi }}\sin \frac{\mathrm{\pi }}{4} \\ \Rightarrow I=1+\frac{4}{\mathrm{\pi }\sqrt{2}} \\ \Rightarrow I=1+\frac{2\sqrt{2}}{\mathrm{\pi }} \end{gathered}$$

Disclaimer: The answer given in the book has some error. The solution here is created according to the question given in the book.